Final answer:
The approximate increase in the density of water at a depth of 400 meters is roughly 2.0794 kg/m³, calculated by using the initial density, depth, gravitational acceleration, and the bulk modulus, which reflects the resistance of water to compression.
Step-by-step explanation:
To calculate the approximate change in density of water at a depth of 400 meters below the surface, we use the formula derived from the definition of the bulk modulus, B, which expresses the substance's resistance to compression. The bulk modulus is given by B = -V(ΔP/ΔV), where ΔP is the change in pressure, V is the initial volume, and ΔV is the change in volume.
In this case, the pressure increase can be found by using the formula P = hpg, where h is the depth, p is the initial density of the water, and g is the acceleration due to gravity.
Since we are only asked for an approximation, we can apply the gravitational acceleration as approximately 9.81 m/s².
First, we find the increase in pressure at the depth of 400 m:
P = hpg = (400 m)(1030 kg/m³)(9.81 m/s²) = 4.0362 × 10&sup6; Pa.
Now we can calculate the change in volume using the bulk modulus:
ΔV/V = -ΔP/B
The volume change is proportional to the change in density (since density is mass over volume, and mass remains constant), so
Δρ/ρ = -ΔP/B
Δρ = -(ρΔP)/B
Δρ = -(1030 kg/m³ × 4.0362 × 10&sup6; Pa) / (2 × 10&sup9; N/m²)
Finally, we find the change in density:
Δρ = -2.0794 kg/m³
The negative sign indicates the density increases, so the approximate change in the density of water at 400 m depth is an increment by roughly 2.0794 kg/m³.